Trying to understand a proof in Rudin concerning winding number

Question

In the proof of theorem 10.10 in Real and complex analysis Rudin states that if we will differentiate $$\phi(t) = \exp \left\{\int_a^t \frac{\gamma'(s)}{\gamma(s)-z} \,\textrm{d}s\right\}, \textrm{we obtain the equality: } \frac{\phi'(t)}{\phi(t)} = \frac{\gamma'(t)}{\gamma(t)-z} $$ which is true almost everywhere (with exception of set $S$, which is finite: $\gamma$ doesn't have to have the derivative everywhere). I can understand this: $$\ln \phi(t) = \int_a^t \frac{\gamma'(s)}{\gamma(s)-z} \,\textrm{d}s \textrm{ implies } (\ln \phi(t))' = \frac{\phi'(t)}{\phi(t)} = \dots = \frac{\gamma'(t)}{\gamma(t) - z}.$$ I hope that the preceeding reasoning is correct (if not, what to do?). But how to conclude that $\phi / (\gamma -z )$ is a continuous function on $[a, b]$ whose derivative is zero on $[a,b] \setminus S$? Here $\gamma \colon [a,b] \to \mathbb C$ is a closed path. Especially the derivative part is a trouble. Of course, it is true that $$\frac{\phi(t)}{\gamma(t)-z} = \frac{\phi'(t)}{\gamma'(t)}.$$ If RHS is constant, then $\phi'(t) = c \gamma'(t)$. Using the definition I write $$\phi'(t) = \exp \left\{\int_a^t \frac{\gamma'(s)}{\gamma(s)-z} \,\textrm{d}s\right\} \cdot \frac{1}{\gamma(t)-z} \cdot \gamma'(t) = c(t) \cdot \gamma'(t).$$ Looking at this I'm not convinced that $c$ is $t$-independent (constant).

Answer 1

It turns out that the answer is quite obvious. We know that $\phi(t) \gamma'(t) = \phi'(t) (\gamma(t)-z)$, so: $$\left[\frac {\phi(t)}{\gamma(t)-z}\right]' = \frac{\phi'(t)(\gamma(t)-z) - \phi (t) \gamma'(t)}{(\gamma(t)-z)^2} = 0.$$

I will leave it here, maybe somebody will find it helpful.